The numbet of values of r for which the coefficients of rth and (r + 1)th terms in the expansion of `(1 + x)^(3n)` are in the ratio 1 :2, is __ _

Answer

To solve the problem, we need to find the number of values of \( r \) for which the coefficients of the \( r \)-th and \( (r + 1) \)-th terms in the expansion of \( (1 + x)^{3n} \) are in the ratio \( 1:2 \).

### Step-by-Step Solution:

1. **Identify the coefficients of the terms**:
The \( r \)-th term in the expansion of \( (1 + x)^{3n} \) is given by:
\[ T_r = \binom{3n}{r} x^r \]
and the \( (r + 1) \)-th term is:
\[ T_{r+1} = \binom{3n}{r + 1} x^{r + 1} \]

2. **Set up the ratio of coefficients**:
We are given that the ratio of the coefficients of the \( r \)-th and \( (r + 1) \)-th terms is \( 1:2 \). Therefore, we can write:
\[ \frac{\binom{3n}{r + 1}}{\binom{3n}{r}} = \frac{1}{2} \]

3. **Use the property of binomial coefficients**:
Using the property of binomial coefficients, we have:
\[ \frac{\binom{3n}{r + 1}}{\binom{3n}{r}} = \frac{3n - r}{r + 1} \]
Thus, we can set up the equation:
\[ \frac{3n - r}{r + 1} = \frac{1}{2} \]

4. **Cross-multiply to eliminate the fraction**:
Cross-multiplying gives us:
\[ 2(3n - r) = r + 1 \]
Expanding this, we get:
\[ 6n - 2r = r + 1 \]

5. **Rearranging the equation**:
Rearranging the equation yields:
\[ 6n - 1 = 3r \]
Therefore, we can express \( r \) as:
\[ r = \frac{6n - 1}{3} \]

6. **Determine the conditions for \( r \)**:
For \( r \) to be an integer, \( 6n - 1 \) must be divisible by \( 3 \). We can check the divisibility:
\[ 6n - 1 \equiv 0 \mod{3} \]
Since \( 6n \equiv 0 \mod{3} \), we have:
\[ -1 \equiv 0 \mod{3} \]
This is not possible, indicating that \( 6n - 1 \) is never divisible by \( 3 \).

7. **Conclusion**:
Since \( r \) cannot be an integer for any integer \( n \), there are no values of \( r \) for which the coefficients of the \( r \)-th and \( (r + 1) \)-th terms are in the ratio \( 1:2 \).

Thus, the number of values of \( r \) is:
\[ \boxed{0} \]